The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives $11.4$ years; the standard deviation is $1.7$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living less than $13.1$ years.
Answer: $11.4$ $9.7$ $13.1$ $8$ $14.8$ $6.3$ $16.5$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $11.4$ years. We know the standard deviation is $1.7$ years, so one standard deviation below the mean is $9.7$ years and one standard deviation above the mean is $13.1$ years. Two standard deviations below the mean is $8$ years and two standard deviations above the mean is $14.8$ years. Three standard deviations below the mean is $6.3$ years and three standard deviations above the mean is $16.5$ years. We are interested in the probability of a meerkat living less than $13.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the meerkats will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the meerkats will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $9.7$ years and the other half $({16\%})$ will live longer than $13.1$ years. The probability of a particular meerkat living less than $13.1$ years is ${68\%} + {16\%}$, or $84\%$.